Derivatives of Inverse Trig Functions
If y = arcsinx, then siny = x
So, cosy = ?
1
x
y
?
1
That matters because if I want to differentiate y = arcsinx, I need to approach it from it's inverse.
Differentiate y = arcsinx
siny = x
cosy(y') = x '
y' =
'
(implicit differentiaton) = Thus, d/dx [arccosx] = ?
.
2
d/dx [arctanx] = d/dx [arccotx] = 3
d/dx [arcsecx] = d/dx [arccscx] = 4
Observations?
5
Find the derivative of f(x) = arcsin(2x)
Find the derivative of f(x) = arctan√x-1
6
Find the derivative of the function
f(x) = cos(arcsin(x2))
7
Find the derivative of the function:
y = x(arcsinx) + √1 - x2
8
9
10
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